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# Physics/Kepler's Laws

Question

Problem
(Given: The value of the universal gravitational constant is 6.672e−11 N · m2/kg2)

Objects with masses of 224 kg and 614 kg are separated by 0.369 m. A 61.1 kg mass is placed midway between them.

a) Find the magnitude of the net gravitational force exerted by the two larger masses on the 61.1 kg mass. Answer in units of N

b) Leaving the distance between the 224 kg and the 614 kg masses fixed, at what distance from the 614 kg mass (other than infinitely remote ones) does the 61.1 kg mass experience a net
force of zero? Answer in units of m

The gravitational force between two bodies is given by:
Fg=G*m1*m2/R^2
In this case the gravitational force of the left mass m1 on the center mass m3 will be:
Fg13=G*m1*m3/R13^2=(6.67x10^-11)*224*61.1/(.369/2)^2=2.68x10^-5N left
While the gravitational force on the center mass by the right mass will be:
Fg23=G*m2*m3/R23^2=2=(6.67x10^-11)*614*61.1/(.369/2)^2=7.35x10^-5 right
Since force is a vector subtract to determine the net force,on th center mass
Fgnet=Fg23-Fg13=4.67x10^-5N right
For the center mass to feel zero net force these two gravitational forces must be equal:
Fg13=Fg23
G*m1*m3/R13^2=G*m2*m3/R3023^2
Simplifying:
m1/R13^2=m2/R23^2
Solve for R23 in terms of R13:
m2/m1=R23^2/R13^2
R23=R13*sqrt(m2/m1)=R13*sqrt(614/224)=1.66*R13
Since these two distance must add up to 0.369m:
R13+R23=R13+1.66*R13=2.66*R13=0.369
Therefore, R13 becomes:
R13=0.369/1.66=0.139m
And:
R23=0.369-R13=0.369-0.139=0.230m

Physics